Rearrangement in proof for Euler's formula

Question

In the proof for Euler's formula, we expand $e^{ix}$ as a Taylor series, rearrange the terms, factor out $i$, and thus obtain the Taylor series for $\sin (x)$ and $\cos(x)$. However, this rearrangement can only be done if the Taylor series for $e^{ix}$ is absolutely convergent, by the Riemann series theorem.

I know how to prove that a series of real terms is absolutely convergent. However, how do you do the same for a series of complex terms, like the one obtained in the Taylor series expansion of $e^{ix}$?

Answer 1

Rearrangements are not the issue here. The basic result is this:

Thm: Suppose $z_n=x_n+iy_n$ is a sequence of complex numbers, where $x_n,y_n$ are the real and imaginary parts of $z_n.$ Then $\sum_{n=1}^{\infty} z_n$ converges in $\mathbb C$ iff both $\sum_{n=1}^{\infty} x_n$ and $\sum_{n=1}^{\infty} y_n$ converge in $\mathbb R.$ In the case of convergence we have

$$\sum_{n=1}^{\infty} z_n = \left(\sum_{n=1}^{\infty} x_n\right )+i\left(\sum_{n=1}^{\infty} y_n\right).$$

This implies Euler's formula for $e^{ix}.$

Answer 2

The complex numbers work the same way $\mathbb{R}^2$ does as a metric spaces but the generalization is the same for $\mathbb{R}^n$, we pass from the point to its length by using the distance function induced by the Euclidean norm. This often requires repeated use of the triangle inequality but the flow of the argument remains essentially the same for the multivariable Taylor series.

Answer 3

First, the series for $e^{it}$ does converge absolutely: $$\sum\frac{|it|^n}{n!} =\sum\frac{|t|^n}{n!}=e^{|t|}<\infty.$$More important, this doesn't matter, because there are no rearrangements in the proof. We just use this:

If $\sum z_{2n}$ and $\sum z_{2n+1}$ both converge, then so does $\sum z_n$, with $\sum z_n=\sum z_{2n}+\sum z_{2n+1}$.

Proof: Define $$a_n=\begin{cases}z_n,&(n\text{ even}), \\0,&n\text{ odd}.\end{cases}$$Let $b_n=z_n-a_n$. Then $z_n=a_n+b_n$, hence $\sum z_n=\sum a_n+\sum b_n$.

The "rearrangements" in Riemann's theorem are sums of the form $$\sum_nz_{\sigma(n)},$$where $\sigma:\Bbb N\to\Bbb N$ is a bijection. Nothing like that here...